What is the molecular formula for a compund that is 82.76% C has a molecular mass of 58.1 grams per mole, and is composed solely of carbon and hydrogen

Molecular formula for the compound is: C₄H₁₀

Explanation:

We analyse the data given.

82.76% means, that in 100g of compound 82.76 g are C

We have the molar mass which is 58.1 g/mol, so let's determine the mass of C, and then, its moles. Therefore we make this rule of three:

In 100 g of compound we have 82.76 g of C

In 58.1 g of compound we may have (85.1 . 82.76) / 100 = 48 g of C

If we convert the mass to moles: 48 g . 1mol / 12g = 4 moles of C

we notice that the compound has 4 moles of C

Molar mass of compound is 58.1 g/mol

48 g of compound are C, therefore (58 - 48) = 10 g are  from H

If we convert the mass to moles → 10 g . 1 mol/1 g = 10 moles

Molecular formula for the compound is: C₄H₁₀

Notice: Centesimal composition for H in the compound

(10g / 58 g) . 100 = 17.24 %

17.24 % + 82.76% = 100%

C4H10.

Explanation:

The  % of Hydrogen = 100 - 82.76 = 17.24%

Dividing  by the atomic masses

C = 82.76 / 12.011 = 6.89

H = 17.24 / 1.008 = 17.1

C: H = 1:2.48

= 2:5

So the empirical formula is C2H5   = 2*12.011+ 5.04 = 29.05 g

Now the molecular mass is 58.1 g which is 2 * 29.05  so the molecular formula is

2 * (C2H5) = C4H10.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the answer is barium and copper bromide